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199. Binary Tree Right Side View

Description

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

 

Example 1:

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Example 2:

Input: root = [1,null,3]
Output: [1,3]

Example 3:

Input: root = []
Output: []

 

Constraints:

  • The number of nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100

Solutions

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode() {}
     *     TreeNode(int val) { this.val = val; }
     *     TreeNode(int val, TreeNode left, TreeNode right) {
     *         this.val = val;
     *         this.left = left;
     *         this.right = right;
     *     }
     * }
     */
    class Solution {
        public List<Integer> rightSideView(TreeNode root) {
            List<Integer> ans = new ArrayList<>();
            if (root == null) {
                return ans;
            }
            Deque<TreeNode> q = new ArrayDeque<>();
            q.offer(root);
            while (!q.isEmpty()) {
                ans.add(q.peekLast().val);
                for (int n = q.size(); n > 0; --n) {
                    TreeNode node = q.poll();
                    if (node.left != null) {
                        q.offer(node.left);
                    }
                    if (node.right != null) {
                        q.offer(node.right);
                    }
                }
            }
            return ans;
        }
    }
    
  • /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
     *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
     * };
     */
    class Solution {
    public:
        vector<int> rightSideView(TreeNode* root) {
            vector<int> ans;
            if (!root) {
                return ans;
            }
            queue<TreeNode*> q{ {root} };
            while (!q.empty()) {
                ans.emplace_back(q.back()->val);
                for (int n = q.size(); n; --n) {
                    TreeNode* node = q.front();
                    q.pop();
                    if (node->left) {
                        q.push(node->left);
                    }
                    if (node->right) {
                        q.push(node->right);
                    }
                }
            }
            return ans;
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode:
    #     def __init__(self, val=0, left=None, right=None):
    #         self.val = val
    #         self.left = left
    #         self.right = right
    class Solution:
        def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
            ans = []
            if root is None:
                return ans
            q = deque([root])
            while q:
                ans.append(q[-1].val) # add last node of previous level traversal results
                for _ in range(len(q)):
                    node = q.popleft()
                    if node.left:
                        q.append(node.left)
                    if node.right:
                        q.append(node.right)
            return ans
    
    #############
    
    # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    from collections import deque
    
    
    class Solution(object):
      def rightSideView(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
    
        def dfs(root, h):
          if root:
            if h == len(ans):
              ans.append(root.val)
              # pre-order, all the way to the right
            dfs(root.right, h + 1)
            dfs(root.left, h + 1)
    
        ans = []
        dfs(root, 0)
        return ans
    
    
  • /**
     * Definition for a binary tree node.
     * type TreeNode struct {
     *     Val int
     *     Left *TreeNode
     *     Right *TreeNode
     * }
     */
    func rightSideView(root *TreeNode) (ans []int) {
    	if root == nil {
    		return
    	}
    	q := []*TreeNode{root}
    	for len(q) > 0 {
    		ans = append(ans, q[len(q)-1].Val)
    		for n := len(q); n > 0; n-- {
    			node := q[0]
    			q = q[1:]
    			if node.Left != nil {
    				q = append(q, node.Left)
    			}
    			if node.Right != nil {
    				q = append(q, node.Right)
    			}
    		}
    	}
    	return
    }
    
  • /**
     * Definition for a binary tree node.
     * class TreeNode {
     *     val: number
     *     left: TreeNode | null
     *     right: TreeNode | null
     *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
     *         this.val = (val===undefined ? 0 : val)
     *         this.left = (left===undefined ? null : left)
     *         this.right = (right===undefined ? null : right)
     *     }
     * }
     */
    
    function rightSideView(root: TreeNode | null): number[] {
        const ans = [];
        if (!root) {
            return ans;
        }
        const q = [root];
        while (q.length) {
            const n = q.length;
            ans.push(q[n - 1].val);
            for (let i = 0; i < n; ++i) {
                const { left, right } = q.shift();
                left && q.push(left);
                right && q.push(right);
            }
        }
        return ans;
    }
    
    
  • // Definition for a binary tree node.
    // #[derive(Debug, PartialEq, Eq)]
    // pub struct TreeNode {
    //   pub val: i32,
    //   pub left: Option<Rc<RefCell<TreeNode>>>,
    //   pub right: Option<Rc<RefCell<TreeNode>>>,
    // }
    //
    // impl TreeNode {
    //   #[inline]
    //   pub fn new(val: i32) -> Self {
    //     TreeNode {
    //       val,
    //       left: None,
    //       right: None
    //     }
    //   }
    // }
    use std::rc::Rc;
    use std::cell::RefCell;
    use std::collections::VecDeque;
    impl Solution {
        pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
            let mut res = vec![];
            if root.is_none() {
                return res;
            }
            let mut q = VecDeque::new();
            q.push_back(root);
            while !q.is_empty() {
                let n = q.len();
                res.push(q[n - 1].as_ref().unwrap().borrow().val);
                for _ in 0..n {
                    if let Some(node) = q.pop_front().unwrap() {
                        let mut node = node.borrow_mut();
                        if node.left.is_some() {
                            q.push_back(node.left.take());
                        }
                        if node.right.is_some() {
                            q.push_back(node.right.take());
                        }
                    }
                }
            }
            res
        }
    }
    
    

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